Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 13.6 due 10/05/2021 at 11:59pm MST
Problem 1. (1 point)
Match the parametric equations with the verbal descriptions of the
surfaces by putting the letter of the verbal description to the left of
the letter of the parametric equation.
1. r(u,v) = ui+cosvj+sinvk
2. r(u,v) = ucosvi+usinvj+u2k
3. r(u,v) = ui+ucosvj+usinvk
4. r(u,v) = ui+vj+ (2u−3v)k
A. cone
B. circular paraboloid
C. circular cylinder
D. plane
Solution:
SOLUTION
1. The corresponding parametric equations for the surface are
x=u,y=cosv,z=sinv. For any point (x,y,z)on the surface,
we have y2+z2=1. With no restritions on the parameters, the
surface is y2+z2=1 which we recognize as a circular cylinder.
Thus the answer is C.
2. The corresponding parametric equations for the surface are
x=ucosv,y=usinv,z=u2. For any point (x,y,z)on the sur-
face, we have x2+y2=z, which we recognize as the equation of
a circular paraboloid. Thus the answer is B.
3. The corresponding parametric equations for the surface are
x=u,y=ucosv,z=usinv. For any point (x,y,z)on the surface,
we have x2=y2+z2, which we recognized as the equation of a
cone. Thus the answer is A.
4. The corresponding parametric equations are x=u,y=v,z=
2u−3v. For any point (x,y,z)on the surface, we have z=2x−3y,
which we recognize as the equation of a plane. Thus the answer
is D.
Answer(s) submitted:
•c
•b
•a
•d
(correct)
Correct Answers:
•C
•B
•A
•D
Problem 2. (1 point)
Consider x=h(y,z)as a parametrized surface in the natural way.
Write the equation of the tangent plane to the surface at the point
(4,−1,−5)given that ∂h
∂y(−1,−5) = 2 and ∂h
∂z(−1,−5) = 3.
.
Solution:
SOLUTION
The natural parametrization of the surface is
r(y,z) = hh(y,z),y,zi.
We have ry(−1,−5) = D∂h
∂y(−1,−5),1,0E=h2,1,0iand
rz(−1,−5) = D∂h
∂z(−1,−5),0,1E=h3,0,1i.
A normal to the tangent plane is given by ry(−1,−5)×
rz(−1,−5) = h2,1,0i×h3,0,1i=h1,−2,−3i.
Thus the equation of the tangent plane is
x−4−2(y+1)−3(z+5) = 0
or, in simplified form,
x−2y−3z=21
Answer(s) submitted:
•x-4-2(y+1)-3(z+5)=0
(correct)
Correct Answers:
•x-4-2*(y--1)-3*(z--5)=0
1
Problem 3. (1 point)
Consider the surface with parametric equations r(s,t) = hst,s+
t,s−ti.
A) Find the equation of the tangent plane at (2,3,1).
.
B) Find the surface area under the restriction s2+t2≤1
Solution:
SOLUTION
A)
rt(s,t) = hs,1,−1iand rs(s,t) = ht,1,1i. Since the point (2,3,1)
corresponds to s=2,t=1, a normal vector to the surface at
(2,3,1)is given by
rs(2,1)×rt(2,1) = h1,1,1i×h2,1,−1i=h−2,3,−1i.
An equation of the tangent plane is
−2(x−2) + 3(y−3)−(z−1) = 0
B)
rs×rt=h−2,s+t,t−siand
|rs×rt|=p4+ (s+t)2+ (t−s)2=√4+2s2+2t2.
The surface area under the given restrictions is then
ZZs2+t2≤1p4+2s2+2t2dA =Z2π
0Z1
0
rp4+2r2dr
=2π1
6(4+2r2)3/21
0
=π
3(6√6−8)
=π2√6−8
3
Answer(s) submitted:
•-2(x-2)+3(y-3)-(z-1)=0
•7.0130
(correct)
Correct Answers:
•-2*(x-2)+3*(y-3)-(z-1)=0
•7.01301755236959
Problem 4. (1 point)
Find the surface area of that part of the plane 10x+5y+z=6 that
lies inside the elliptic cylinder x2
100 +y2
4=1
Surface Area =
Solution:
SOLUTION
We have z=f(x,y) = 6−10x+5yand Dis the elliptic disk
x2
100 +y2
4≤1, so by Formula 9, the area of the surface is
A(S) = RRDr1+∂z
∂x2+∂z
∂y2
dA
=RRDp1+ (−10)2+ (−5)2dA
=√126RRDdA
=√126 ·Area(D)
=√126π(10)(2)
=20√126π
Answer(s) submitted:
•sqrt(126)(10)(2)pi
(correct)
Correct Answers:
•705.28580151234
2
Problem 5. (1 point)
Write down the iterated integral which expresses the surface area
of z=y6cos6xover the triangle with vertices (−1,1),(1,1),(0,2):
Zb
aZg(y)
f(y)ph(x,y)dxdy
a=
b=
f(y) =
g(y) =
h(x,y) =
Solution:
SOLUTION
The line between the points (−1,1)and (0,2)has equation
x=y−2, while the line between the points (1,1)and (0,2)
has equation x=2−y. Thus the region of integration, D, is
D={(x,y)|1≤y≤2,y−2≤x≤2−y}.
The surface area is
A(S) = ZZDs1+∂z
∂x2
+∂z
∂y2
dA
=Z2
1Z2−y
y−2q1+ (−6y6cos5xsinx)2+ (6y5cos6x)2dxdy
Answer(s) submitted:
•1
•2
•y-2
•
•
(score 0.6)
Correct Answers:
•1
•2
•y-2
•2-y
•1 + y**(2*6) * 6**2 * cos(x)**(2*6-2)* sin(x)**2 + 6**2 * y**(2*6-2) * cos(x)**(2*6)
Problem 6. (1 point)
The vector equation r(u,v) = ucosvi+usinvj+vk, 0 ≤v≤9π,
0≤u≤1, describes a helicoid (spiral ramp). What is the surface
area?
Solution:
SOLUTION
ru=cosvi+sinvj,rv=−usinvi+ucosvj+kand ru×rv=
sinvi−cosvj+uk.
Then |ru×rv|=psin2v+cos2v+u2=√1+u2and the surface
area is given by
A(S) = Z9π
0Z1
0p1+u2dudv
=Z9π
0
dvZ1
0p1+u2du
=9π
2(√2+ln(1+√2))
.
Answer(s) submitted:
•32.453
(correct)
Correct Answers:
•32.4530987589317
3
Problem 7. (1 point)
Find the surface area of the part of the sphere x2+y2+z2=9 that
lies above the cone z=px2+y2
Solution:
SOLUTION
Since the cone intersects the sphere in the circle x2+y2=9
2,z=
q9
2and we want the portion of the sphere above this, we can
parametrize the surface as x=x,y=y,z=p9−x2−y2where
x2+y2≤9
2. The surface area is
A(S) = RRx2+y2≤9
2r1+∂z
∂x2+∂z
∂y2
dA
=RRx2+y2≤9
2s1+−2x
2√9−x2−y22
+−2y
2√9−x2−y22
dA
=RRx2+y2≤9
2q1+x2
9−x2−y2+y2
9−x2−y2dA
=RRx2+y2≤9
2q9
9−x2−y2dA
=3Rq9
2
0R2π
0
r
√9−r2dθdr
=6πRq9
2
0
r
√9−r2dr
Using the substitution u=9−r2,du =−2rdr, yields
A(S) = −3πR
9
2
9u−1/2du
=3π[2√u]9
9
2
=6πh3−3
√2i
=18π1−1
√2
Answer(s) submitted:
•16.5627
(correct)
Correct Answers:
•16.562721321191
Problem 8. (1 point)
Find the area cut out of the cylinder x2+z2=100 by the cylinder
x2+y2=100.
Solution:
SOLUTION
We first find the area of the face of the surface that intersects the
positive zaxis. Let S1be this surface. A parametric representation
of the surface is x=x,y=y,z=√100 −x2with x2+y2≤100.
Then
A(S1) = RRx2+y2≤100 r1+∂z
∂x2+∂z
∂y2
dA
=RRx2+y2≤100 s1+−2x
√100−x22
+02dA
=RRx2+y2≤100 q1+x2
100−x2dA
=RRx2+y2≤100
√100
√100−x2dA
=10R10
−10 R√100−x2
−√100−x2
1
√100−x2dydx
=4(10)R10
0R√100−x2
0
1
√100−x2dydx (by the symmetry of the surface)
This integral is improper when x=10, so
A(S1) = 40limt→10−Rt
0R√100−x2
0
1
√100−x2dydx
=40limt→10−Rt
0
√100−x2
√100−x2dx
=40limt→10−Rt
0dx
=40limt→10−t
=400.
Since the complete surface consists of two congruent faces, the
total surface is A(S) = 2·400 =800.
Answer(s) submitted:
•180
(incorrect)
Correct Answers:
•800
4
Problem 9. (1 point)
If a parametric surface given by r1(u,v) = f(u,v)i+g(u,v)j+
h(u,v)kand −1≤u≤1,−1≤v≤1, has surface area equal to
2, what is the surface area of the parametric surface given by
r2(u,v) = 3r1(u,v)with −1≤u≤1,−1≤v≤1?
Solution:
SOLUTION
Let S2be the surface parametrized by r2and S1the surface
parametrized by r1. We have, ∂r2
∂u=3∂r1
∂u,∂r2
∂v=3∂r1
∂v,and
∂r2
∂u×∂r2
∂v
=32
∂r1
∂u×∂r1
∂v
.
Thus
A(S2) = Z1
−1Z1
−1
∂r2
∂u×∂r2
∂v
dvdu
=9Z1
−1Z1
−1
∂r1
∂u×∂r1
∂v
dudv
=9·A(S1)
=9(2) = 18
Answer(s) submitted:
•9
(incorrect)
Correct Answers:
•18
Problem 10. (1 point)
Parameterize the plane through the point (−3,4,2)with the nor-
mal vector h4,−5,−1i
~r(s,t) =
(Use s and t for the parameters in your parameterization, and en-
ter your vector as a single vector, with angle brackets: e.g., as
¡1+s+t,s-t,3-t¿.)
Solution:
SOLUTION
To parameterize the plane we need two nonparallel vectors~v1and
~v2that are parallel to the plane. Such vectors are perpendicular to
the normal vector to the plane, ~n=h4,−5,−1i. We can choose
any vectors ~v1and ~v2such that ~v1·~n=~v2·~n=0.
One choice is
~v1=h5,4,0i, ~v2=h1,0,4i.
Letting~r0=h−3,4,2i, we have the parameterization
~r(s,t) =~r0+s~v1+t~v2=h5s−3+t,4+4s,2+4ti.
Answer(s) submitted:
•<-3+s-t,4-4s,2+4t>
(incorrect)
Correct Answers:
•<5*s-3+t,4+4*s,2+4*t>
5
Problem 11. (1 point)
For a sphere parameterized using the spherical coordinates θand
φ, describe in words the part of the sphere given by the restrictions
π/4≤θ≤π/2 0 ≤φ≤π/2
and
π/2≤θ≤π0≤φ≤π.
Then pick the figures below that match the surfaces you described.
π/4≤θ≤π/2 0 ≤φ≤π/2 : [?/1/2/3/4/5/6/7/8]
π/2≤θ≤π0≤φ≤π: [?/1/2/3/4/5/6/7/8]
(Click on any graph to see a larger version.)
1. 2. 3. 4.
5. 6. 7. 8.
Solution:
SOLUTION
The restrictions π/4≤θ≤π/2 0 ≤φ≤π/2 : correspond to 4.
The restrictions π/2≤θ≤π0≤φ≤π: correspond to 6.
Answer(s) submitted:
•4
•6
(correct)
Correct Answers:
•4
•6
Problem 12. (1 point)
Consider the cone shown below.
If the height of the cone is 8 and the base radius is 9, write a pa-
rameterization of the cone in terms of r=sand θ=t.
x(s,t) = ,
y(s,t) = , and
z(s,t) = , with
≤s≤and
≤t≤.
Solution:
SOLUTION
Since the parameterization is specified to be in terms of the radius
rand angle θ, we find x,yand zin terms of the parameters r=s
and θ=t. We have
x=scost,
y=ssint,
and
z=8(1−1
9s),
with
0≤s≤9 and 0 ≤t≤2π.
There are, of course, other parameterizations, but they will not be
in terms of rand θ, as required in this problem.
Answer(s) submitted:
•scost
•ssint
•8-(8/(9s))
•0
•9
•0
•2pi
(score 0.857143)
Correct Answers:
•s*cos(t)
•s*sin(t)
•8-8/9*s
•0
•9
•0
•2*pi
6
Generated by ©WeBWorK, http://webwork.maa.org, Mathematical Association of America
7
